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Current time:0:00Total duration:8:57

AP Chem: SPQ‑1 (EU), SPQ‑1.A (LO), SPQ‑1.A.1 (EK), SPQ‑1.A.2 (EK), SPQ‑1.A.3 (EK)

- [Instructor] In a previous video, we introduced ourselves to the
idea of average atomic mass, which we began to realize
could be a very useful way of thinking about a
mass at an atomic level, or at a molecular level. But, what we're gonna do in this video is connect it to the masses
that we might actually see in a chemistry lab. You're very unlikely to just
be dealing with one atom, or just a few atoms, or
just a few molecules. You're more likely to
deal with several grams of an actual substance. So, how do we go from the
masses at an atomic scale to the masses, masses of samples that you see in an actual chemistry lab, or in, I guess you could
say, r-scale of the world. Well, the chemistry
community has come up with a useful tool. They said, all right, let's
think about a given element. So, say, lithium. We know its average atomic mass is 6.94, 6.94 unified atomic mass units
per atom, atom of lithium. What if there were a certain
number of atoms of lithium such that if I have that number, so times certain, certain number of atoms, then I will actually
end up with 6.94 grams, grams of lithium. And, this number of atoms is 6.02214076 times 10 to the 23rd power. So, if you have a sample with
this number of lithium atoms, that sample is going to
have a mass of 6.94 grams. Whatever its average atomic mass is in terms of unified atomic mass units, if you have that number of the atom, you will have a mass of that
same number in terms of grams. Now, you might be saying, is
there a name for this number, and there is indeed a name, and it is called Avogadro's number, named in honor of the early
19th century Italian chemist, Amedeo Avogadro. And, in most contexts,
because you're not normally dealing with data with this
many significant digits, we will usually approximate it as 6.022 times 10 to the 23rd power. Now, there's another word
that it's very useful to familiarize yourself with in chemistry, and that's the idea of a mole. Now, what is a mole? It is not a little mark on your cheek. It is not a burrowing animal. Actually, it is both of those things, but, in a chemistry context, a mole is just saying you
have this much of something. The word mole was first
used by the German chemist Wilhelm Ostwald at the
end of the 19th century, and he came up with the word because of its relation to molecule. Now, what does that mean? Well, think about the word dozen. If I say I've got a dozen of eggs, how many eggs do I have? Well, if I have a dozen of eggs, that means I have 12 eggs. So, if I say I have a
mole of lithium atoms, how many lithium atoms do I have? That means that I have 6.02214076 times 10 the to 23rd lithium atoms. Exact same idea, it's just
that Avogadro's number is much hairier of a number than a dozen. So, let's use our new
found powers of the mole and Avogadro's number to start
doing some useful things. Let's say that someone were
to walk up to you and say, hey, you, I have a 15.4
milligram sample of germanium. How many atoms of germanium
am I dealing with? Pause this video and
try to think about that. So, let me clear out some space the periodic table of
elements was taking up. All right, so we started off with 15.4 milligrams of germanium. The first step might be
hey, let's convert this to grams of germanium. And so, we can do a little
bit of dimensional analysis. We can just multiply this, for every one gram of germanium that is equivalent to 1,000 milligrams, milligrams of germanium. And so, if you essentially
multiply by one thousandth or divide by 1,000, we're gonna
get the grams of germanium. And, you can see that in
the dimensional analysis by seeing that that is going
to cancel out with that leaving us with just
the grams of germanium. And, now that we have an
expression for grams of germanium, we can think about moles of germanium. So, how do we do that? Well, we're going to
multiply by some quantity, and in the denominator we're going to want grams of germanium for the
dimensional analysis to work out, grams of germanium, and
in the numerator we want the new expression to be in
terms of moles of germanium. So, one mole of germanium is equal to how many grams of germanium? Well, we see it right over here. Germanium's molar mass
is 72.63 grams per mole. So, for every mole, we have
72.63 grams of germanium. And, you can see that the units work out. These grams of germanium
are going to cancel with the grams of germanium just leaving us with moles of germanium. In an actual chemistry practice, finding out the moles of a substance might actually be the most useful thing, but if you wanted to find out
the actual atoms of germanium that we're dealing with,
we will just multiply by the number of atoms you have per mole. And, this is going to
be true for any element. For every mole, you have
Avogadro's number of atoms. And, we're going to approximate that as 6.022 times 10 to the 23rd
atoms, atoms of germanium, for every one mole, mole of germanium. And so, just to review what we just did, we had milligrams of germanium. You multiply these two together, you'll have grams of
germanium, which makes sense, you're essentially just dividing by 1,000. If you were to multiply
your grams of germanium times the moles per gram, which is really just the reciprocal of this molar mass we got here, and just to make sure
where it makes sense, the units work out nice with
the dimensional analysis, this right over here tells you your moles, moles of germanium. And then, if you take your moles and then you multiply
it by Avogadro's number, it tells you how many
atoms of germanium we have, and that makes sense. If I told you I had a certain
number of dozen of eggs, if I wanted to know how many eggs that is I would multiply by 12. So, this whole expression
is the number of atoms, atoms of germanium. So, we have 15.4 milligrams. If we wanna figure out
how many grams we have, we then divide by 1,000, that's what our dimensional
analysis tells us, and it also makes logical sense, divided by 1,000. So, this is how many grams we have. And then, if we wanna
figure out how many moles, and it's going to be a
small fraction of a mole because a mole is 72.63 grams per mole, we have a small fraction of a gram, much less 72.63 grams. And so, we saw from our analysis to figure out the number of moles, we're now going to
essentially divide by 72.63, so divided by 72.63 is equal to, this is the number of
moles of germanium we have. And, if we wanna figure
out the number of atoms of germanium, we'll then multiply that times Avogadro's number. So, times 6.022 times 10 to the 23rd, and this EE button means times-10-to-the, EE 23rd power, so that's how
you do it on a calculator. And then, that gives us this many atoms. And, let's see, just to get
our significant digits here, our significant figures,
out of all of the things we multiplied, see we had
four significant digits here, four significant digits here, but we only had three over here, so I'm going to round to
three significant digits. So, I'll go to 1.28 times
10 to the 20th atoms. So, we have approximately 1.28 times 10 to the
20th atoms of germanium, which is a lot.